Question Number 131624 by mr W last updated on 06/Feb/21
![](https://www.tinkutara.com/question/18536.png)
Commented by mr W last updated on 06/Feb/21
![an old unsolved question](https://www.tinkutara.com/question/Q131625.png)
$${an}\:{old}\:{unsolved}\:{question} \\ $$
Commented by MJS_new last updated on 07/Feb/21
![I′ll try...](https://www.tinkutara.com/question/Q131676.png)
$$\mathrm{I}'\mathrm{ll}\:\mathrm{try}… \\ $$
Commented by MJS_new last updated on 08/Feb/21
![I get blue area=4A α is the rotation angle of the red square ⇒ area of the left square is (1+8sin^2 α)A area of the right square is (1+8cos^2 α)A](https://www.tinkutara.com/question/Q131787.png)
$$\mathrm{I}\:\mathrm{get}\:\mathrm{blue}\:\mathrm{area}=\mathrm{4}{A} \\ $$$$\alpha\:\mathrm{is}\:\mathrm{the}\:\mathrm{rotation}\:\mathrm{angle}\:\mathrm{of}\:\mathrm{the}\:\mathrm{red}\:\mathrm{square} \\ $$$$\Rightarrow \\ $$$$\mathrm{area}\:\mathrm{of}\:\mathrm{the}\:\mathrm{left}\:\mathrm{square}\:\mathrm{is}\:\left(\mathrm{1}+\mathrm{8sin}^{\mathrm{2}} \:\alpha\right){A} \\ $$$$\mathrm{area}\:\mathrm{of}\:\mathrm{the}\:\mathrm{right}\:\mathrm{square}\:\mathrm{is}\:\left(\mathrm{1}+\mathrm{8cos}^{\mathrm{2}} \:\alpha\right){A} \\ $$
Commented by mr W last updated on 08/Feb/21
![thanks sir!](https://www.tinkutara.com/question/Q131803.png)
$${thanks}\:{sir}! \\ $$